Finding the sum and difference of two drag coefficients

AI Thread Summary
The discussion focuses on calculating the sum and difference of drag coefficients for two bottles subjected to wind forces. The user has determined a sum of drag coefficients as 2.18 but is uncertain about incorporating the lift load cell forces into the torque calculations. There is confusion regarding the terminology of "lift" since the forces are acting horizontally, leading to a debate about the nature of the forces measured by the load cells. The group realizes that the two forces act in opposite directions, which affects the torque calculations. Clarification is sought on how to properly apply the forces to determine the drag coefficients accurately.
Lunia Oriol
Messages
3
Reaction score
0

Homework Statement


I have to determine the sum and difference of the two coefficients of drag between two individual items that make a torque AUT. I'm given the diameter of each bottle (6.35 cm), wind velocity (14.3 m/s), the forces of the lift load cells (+0.0741 N, -0.0741 N), and the force of the drag load cell (0.846 N). Here is an image for clarification.

https://www.dropbox.com/s/k362escpst0m3f0/Photo Jan 12, 22 05 05.jpg?dl=0

Homework Equations



c2a53b7e1a1a64041f3dfca19baf22b2ec0d5c0e


where:

Fd is the drag force, which is by definition the force component in the direction of the flow velocity
ρ
a1d651c28959a0f15127c097ff4488b123d9e708
is the mass density of the fluid,
u
880f91e25cd451d89d1f6d0d06852b56a7b74a32
is the flow speed of the object relative to the fluid,
A
c6aaf5ce10d6add44b973e28fb3d95f37abf3721
is the reference area.

The Attempt at a Solution


I’m pretty sure i found the sum of the drag coefficients since I know the force of the load cell, the velcity, cross area, and density of air (my drag coefficient sum was 2.18), but I don’t know if I’m supposed to now used the force of the lift load cell since the torque is the radius times the perpendicular force. Thanks in advance.

Edited for clarity.
 
Last edited:
Physics news on Phys.org
Lunia Oriol said:
the diameter between the two bottles
Diameter of the bottles?
Lunia Oriol said:
the lift load cells
Lift? Isn't the table horizontal? Some of the text is truncated in your image.
Lunia Oriol said:
if I’m supposed to now used the force of the lift load cell
Yes, of course. What equation can you write?
 
haruspex said:
Diameter of the bottles?

Lift? Isn't the table horizontal? Some of the text is truncated in your image.

Yes, of course. What equation can you write?


I re-uploaded an image that displays the full problem. I determined that there were two load cells that measured lift because the two drag forces pointing to the right are normal to the lift load cells.​

This is where me and my group got lost. I know that torque is a measurement of the radius times the force times sin(theta). If I were to use the force of one load cell (+0.0741), the distance from the middle of the bottle to the sting, and the angle of the force (sin90), I could certainly find the torque. I also know that torque equals the radius times (force2 - force1). But we couldn't go on from there. Initially we thought that because the lift load cells had essentially the same magnitude of force, the difference of the drag coefficients would be zero because we would be plugging in the same values to find the coefficients of drag, but it didn't seem plausible.
 
Lunia Oriol said:
two load cells that measured lift
I still do not understand why you call it lift. That implies vertical force, no? The top diagram is the view from above, so all the forces are in the horizontal plane.
Lunia Oriol said:
the lift load cells had essentially the same magnitude of force
But what are the signs of the two torques? Do they both act the same way about the sting?
 
Lunia Oriol said:
This is where me and my group got lost. I know that torque is a measurement of the radius times the force times sin(theta). If I were to use the force of one load cell (+0.0741), the distance from the middle of the bottle to the sting, and the angle of the force (sin90), I could certainly find the torque. I also know that torque equals the radius times (force2 - force1). But we couldn't go on from there. Initially we thought that because the lift load cells had essentially the same magnitude of force, the difference of the drag coefficients would be zero because we would be plugging in the same values to find the coefficients of drag, but it didn't seem plausible.
haruspex said:
I still do not understand why you call it lift. That implies vertical force, no? The top diagram is the view from above, so all the forces are in the horizontal plane.

But what are the signs of the two torques? Do they both act the same way about the sting?

No, they act in opposite directions. If it's not lift, then what's the force LC1 and LC2 are measuring?
 
Lunia Oriol said:
No, they act in opposite directions.
How do you get that? The two forces are of opposite sign and are on opposite sides of the axis.
Lunia Oriol said:
If it's not lift, then what's the force LC1 and LC2 are measuring?
The three FLC forces are presumably supplied by apparatus to hold the platform steady. The values would be measured by strain gauges.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top